For logarithms, there is the change of base formula $\log_a x = \frac{\log_b x}{\log_b a}$. So, all logarithms are constant multiples of each other, because the logarithm in base $a$ can be expressed as $\frac{1}{\log_b a}$ times the logarithm in base $b$.
Now, it seems the corresponding result isn’t true for exponential functions? Is there an intuitive explanation for why not? Intuitively, one might reason “the exponential is just the inverse of the logarithm, and so if logarithms are within multiples of each other, so are exponentials”. What is wrong with this reasoning?
We can ask the general question: If a family of invertible functions are within, say, constant multiples of each other (or some other relation), can we say anything about a corresponding relationship within the family of inverses?
My attempt:
I claim that exponentials are not within constant multiples of each other. Indeed, suppose there are two bases $a, b$ and a constant $k$ such that $a^x = kb^x$ for all $x$. Solving for $k$ gives $\displaystyle k = \frac{a^x}{b^x} = \frac{e^{x \ln a}}{e^{x \ln b}} = e^{x (\ln a – \ln b)}$. But this is not a constant function.
Now, this derivation shows we can multiply the variable $x$ by the constant $\ln a – \ln b$, so in some sense there is still “multiplication by a constant”, but I don’t really understand why we multiply the argument $x$ by a constant, whereas for logarithms it is just a simple universal constant.
Best Answer
Suppose you have two functions $f(x)$ and $g(x)$ which are both invertible, and they differ by multiplication by a constant:
$$ g(x) = k \cdot f(x) \tag{for all $x$} $$
Then if the point $(x,y)$ is on the graph of $f$, then the point $(x,ky)$ is on the graph of $g$.
If $f$ is a function and $(x,y)$ is on its graph, then $(y,x)$ is on the graph of $f^{-1}$. So in the example above, $(y,x)$ is on the graph of $f^{-1}$ and $(ky,x)$ is on the graph of $g^{-1}$.
So comparing the graphs of $f^{-1}$ and $g^{-1}$, points with the same $y$-coordinate have their $x$-coordinates differ by a constant multiple. This is what you are seeing the the arguments of the exponential functions differing by a constant multiple (rather than the functions themselves).
Intuitively, passing from a function to its inverse switches the roles of the $x$ and $y$ coordinates on the graph, so the roles of inputs and outputs of the function are changed. For logarithms, the outputs differ by a constant multiple, so for exponentials, the inputs differ by constant multiple.